Angular and CoM Velocities of a Solid Sphere

AI Thread Summary
The discussion focuses on calculating the angular velocity (ω2) and center of mass velocity (vcm) of a solid sphere rolling down a curved rail from height h1 to h2. The kinetic energy equations used include both translational and rotational components, with the moment of inertia for a solid sphere being I = (2/5)mR^2. The derived equations show that the angular velocity is ω = √((10/7)g(h1-h2))/R and the center of mass velocity is v = √((10/7)g(h1-h2)). The solution confirms that v and ω are related by the equation v = ωR, indicating a correct approach to the problem. The calculations appear accurate and consistent with the physics principles involved.
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Homework Statement


A solid sphere of mass M and radius R is rolling,without slipping, down a curved rail. The sphere is initially at rest at a height of h1. Find the angular velocity ω2 and the center of mass velocity of the sphere vcm at the end of the rail of height h2. You may assume that no vibration and heat are generated as the sphere rolls along the rail.

Homework Equations


solid sphere, I = \frac{2}{5}mR^2


The Attempt at a Solution


I'm not sure if I began with the correct equation.

KE = \frac{1}{2} m v^2 + \frac{1}{2} I ω^2
= \frac{1}{2} m(ωR)^2 + \frac{1}{2} (\frac{2}{5}mR^2) ω^2
mg(h_1 -h_2) = \frac{7}{10}m ω^2 R^2
ω = √(\frac{10}{7}g(h_1-h_2)) / R

KE = \frac{1}{2} m v^2 + \frac{1}{2} I ω^2
= \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5}mR^2) (\frac{v}{R})^2
mg(h_1 -h_2) = \frac{7}{10}mv^2
v = √(\frac{10}{7}g(h_1-h_2))

Thanks in advance!
 
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Looks good to me. (Of course there was no need to solve it twice, since v = ωR.)
 
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